What does $X^{-1}.Y.X = Z$ represents? How we can further simplify it? I am trying to solve chain matrices multiplication. Where I encounter a problem, which sees like dead end to me.
How further I can decompose $X^{-1} . Y . X = Z$ or simplify it to find matrix $X$ from this multiplication.
Where matrices $X,Y$and $Z$ are 2x2 non-singular matrices.
Any help will be appreciated.
 A: Note that your equation is quadratic in $X$ and therefore the solution is not unique, and not even guaranteed to exist. For example, if $X$ is a solution, then so is $-X$. Depending on the structure of $Y$ and $Z$, you might have more. 
Also, the matrix $Z$ is similar to the matrix $Y$. That is, they both represent the same linear transformation but in different bases. Therefore, your equation does not have a solution if, $\operatorname{tr}(A)\ne\operatorname{tr}(B)$ or $\operatorname{det}(A)\ne\operatorname{det}(B)$.
Otherwise there exist a solution, but as I said, it is not unique. You can easily find it by solving, enrtywise, the 4 equations.
A: Rewriting your equation as $YX = XZ$ you get a special version of a Sylvester equation. An easy solution is to vectorize this equation using the Kronecker product
$$\mathrm{vec}(AYB) = (B^T \oplus A)\mathrm{vec}(Y)$$
(where $\oplus$ shall denote the Kronecker product!). Then,
$$\mathrm{vec}(YX) = (I \oplus Y) \mathrm{vec}(X) = (Z^T \oplus I) \mathrm{vec}(X) = \mathrm{vec}(XZ)$$
and you only have to solve the simple linear equation
$$((I \oplus Y) - (Z^T \oplus I)) \mathrm{vec}(X) = 0.$$
Cheers,
Wieland
